# Ceva’s theorem

Let $ABC$ be a given triangle and $P$ any point of the plane. If $X$ is the intersection point of $AP$ with $BC$, $Y$ the intersection point of $BP$ with $CA$ and $Z$ is the intersection point of $CP$ with $AB$, then

 $\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.$

Conversely, if $X,Y,Z$ are points on $BC,CA,AB$ respectively, and if

 $\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$

then $AX,BY,CZ$ are concurrent.

Remarks: All the segments are directed segments (that is $AB=-BA$), and so theorem is valid even if the points $X,Y,Z$ are in the prolongations (even at the infinity) and $P$ is any point on the plane (or at the infinity).

 Title Ceva’s theorem Canonical name CevasTheorem Date of creation 2013-03-22 11:57:15 Last modified on 2013-03-22 11:57:15 Owner drini (3) Last modified by drini (3) Numerical id 16 Author drini (3) Entry type Theorem Classification msc 51A05 Related topic Triangle Related topic Median Related topic Centroid Related topic Orthocenter Related topic OrthicTriangle Related topic Cevian Related topic Incenter Related topic GergonnePoint Related topic MenelausTheorem Related topic ProofOfVanAubelTheorem Related topic VanAubelTheorem Related topic BisectorsTheorem Related topic DirectedSegment